IP AMaths Notes (Upper Sec, Year 3-4): 16) Applications of Differentiation

Apply derivatives to describe gradient, optimise functions, and connect rates.

Common tasks

• Tangent gradient at \( x = a \): evaluate \( y' \) at \( a \).
• Normal gradient: \( m_\text{normal} = -\dfrac{1}{m_\text{tangent}} \) for non-zero tangents.
• Stationary points satisfy \( y' = 0 \); use second derivative or sign chart for classification.
• Related rates: link quantities via differentiation with respect to time.

Worked example 1 — Tangent and normal

For \( y = x^3 - 3x + 2 \) find the tangent and normal at \( x = 1 \).

1. Derivative: \( y' = 3x^2 - 3 \).
2. Gradient at \( x = 1 \): \( m_t = 0 \) (horizontal tangent).
3. Point: \( y = 1 - 3 + 2 = 0 \) so coordinate \( (1, 0) \).
4. Tangent: \( y = 0 \).
5. Normal would require reciprocal gradient but \( m_t = 0 \) ⇒ normal is vertical line \( x = 1 \).

Worked example 2 — Optimisation

Find the minimum value of \( y = x + \dfrac{9}{x} \) for \( x > 0 \).

1. Derivative: \( y' = 1 - \dfrac{9}{x^2} \).
2. Set \( y' = 0 \): \( 1 = \dfrac{9}{x^2} \) → \( x^2 = 9 \) → \( x = 3 \) (positive regime).
3. Second derivative: \( y'' = \dfrac{18}{x^3} > 0 \) for \( x > 0 \) ⇒ minimum.
4. Minimum value: \( y = 3 + \dfrac{9}{3} = 6 \).

Try this

Water drains from a conical tank (radius \( 2 \) m at the top, height \( 3 \) m). Express rate of change of height in terms of the volume decrease rate \( \dfrac{\mathrm{d}V}{\mathrm{d}t} \).